Balance the equation: NaOH + H$_2$SO$_4$ $\longrightarrow$ Na$_2$SO$_4$ + H$_2$O. The correct set of coefficients (NaOH, H$_2$SO$_4$, Na$_2$SO$_4$, H$_2$O) is:

1. Start with Na: the product Na$_2$SO$_4$ has 2 Na, so place a factor 2 before NaOH. 2. Equation becomes 2NaOH + H$_2$SO$_4$ $\longrightarrow$ Na$_2$SO$_4$ + H$_2$O. 3. Now count H on the left: 2 (from 2NaOH) + 2 (from H$_2$SO$_4$) = 4; place factor 2 before H$_2$O to give 4 H. 4. Check O: left = 2 + 4 = 6, right = 4 (Na$_2$SO$_4$) + 2 (2H$_2$O) = 6. S = 1 each side. Balanced as 2, 1, 1, 2. 5. The set 1,1,1,1 leaves Na and H unequal; the sets beginning 2,2 and 1,2 over-count and unbalance the equation. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 3 "Chemical Reactions and Equations", p.44_